This Calc problem is way too hard...?

Question

Find the
-intervals during which the function is concave up and concave down.
-The x-coordinate for the inflection point
-the absolute and minimum x values
for the function:
f(x)=((5x-8)^5)^1/3 defined on the interval [-5, 5]

Answer 1

First, do a little house-cleaning. You have
f(u) = u^(5/3) where u= 5x-8 and du/dx=5.
So your first derivitive will be d(f[u])/du * du/dx
You can do this once more for the 2d derivitive.
Now you can do your analyses.

Answer 2

ok i attempted this and could get you a start but finishing this one is insane. I got the first derivative to be 25/3((5x-8)^9)^-2/3 So the critical point would be 8/5.
The second derivative would be -250((5x-8)^17)^-5/3
so you need to set the second derivative = 0 to find inflection point. If you do that x sould = 8/5 to make it 0. I can not figure out the concavity though because the function is so hard to read. hope that helps at all.